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When you have a capacitor, current flows even though the "circuit" is not complete. This is because it's possible for electrons to bunch up - temporarily - in a conductor and generate a corresponding electric field. That is what happens in an antenna. An antenna is really a combination of an inductor (a straight wire) and a capacitor (when you put a net ...


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I think that this is a very good question because it makes one think beyond a "standard" explanation. When you study electromagnetic induction you learn about the magnetic flux change through a close loop which produces an induced emf. However the loop does not have to be a conducting loop. If it was an ideal dynamo with no resistance, friction etc no ...


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An electromagnet has a lot of turns of wire, which gives it a property called inductance : the property of resisting a change in the current. You may be familiar with Lenz's Law $$V=-L\frac{d\Phi}{dt}$$ which says that an inductor with inductance $L$ will generate a back e.m.f. $V$ when the flux $\Phi$ through it changes. Now if you drive a current through ...


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You have two basic options: Realize that this is actually just two resistors in series with three parallel resistors, and analyze it using the equivalent resistances of resistors in parallel and series, or Use Kirchoff's laws to derive the equivalent resistance. If you choose option 1, I'll help you out by revealing the parallel resistors in this ...


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The helicopter and the power lines are at different potentials, the difference being so great as to cause the air in between to become a conductor. If you applied such a potential difference across a line worker it would probably result in death. You will note that the line worker is holding a metal stake which has a "pointed" end. This increases the ...


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I had read that resistances only lose energy in the form of heat. Correct. Is that what the authors are talking about? Yes, the author is just using confusing language. Do they mean to say that this internal energy is first stored in the resistor in the form of internal energy and then dissipated as heat? The author means that electrical ...


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In order to make the problem symmetric, consider this: What happens, when you take the dodecahedron and drive current $I$ into it from vertex $A$ and drive $I/20$ out from all every vertex (including $A$)? By Kirchhoff's laws and symmetry, there is a current $I_{A-out} = \frac{(I-I/20)}{3} = \frac{19}{60}I$ going from $A$ to neighbouring vertexes. Now ...


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A capacitor can contain a certain amount of charge for a given voltage: $$Q = CV$$ When you have more than one capacitor in parallel, they have the same voltage (because they are in parallel), and each stores a certain charge. The total charge (at a given voltage) will be the sum of the charges on all the capacitors. Now if you have a certain load (for ...


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The powerline can supply large amount of current, whereas the doorknob cannot. Each power supply comes with voltage and max rated current. Let us say body resistance is 5M ohm. Then current flowing through the body when you touch powerline is 20k/5M = 4mA. Now powerline can easily supply 4mA of current. But the doorknob cannot supply 4mA. So the actual ...


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The Key to your question is the distinguation between ideal and real sources of voltage and current. If we stay at the batteries: At the moment you attach batteries paralell to your body, what happens is that one part of the body has another electric potential than the other one, thus (Ohms law) a current will flow in your body. At this point the current ...


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The current in a circuit is a collective phenomenon from zillions of electrons. It appears due to conductivity, another collective phenomenon . It is a cumulative behavior of atoms and electrons in matter. In insulators, electrons occupy energy levels and have to be actively kicked out of them, with the energy provided by an interaction. Insulators can be ...


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Even if you put a superconductor across the terminals of a voltage source the current would be finite as all real voltage sources have a resistance. A circuit with a voltage source with no resistance does not exist.


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You've asked some really good questions here. Before starting, I want to first mention that the traditional picture of particles moving through a wire in electostatics is missing some physics; for instance, it ignores the quantum mechanical nature of electrons. The reason we still teach this model is because it captures the main effects (the phenomenon of ...


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The fundamental problem is that the Ohm's law is a simple mathematical approximation, neglecting the V-A characteristic of a real source, resistance of wires, or even nonlinear physical response of materials under extreme conditions. Using this approximation outside of its scope leads to a "0/0" problem, which is obviously wrong. Some other physical ...


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You can confine an electric field in a Faraday cage, if you want to, then the volume is finite. – CuriousOne


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When you analyze a circuit, you have to pick a convention for your potential and current flow. It doesn't matter whether you pick left-right or right-left, up-down or down-up, you just have to pick. After that, you do your calculations. If you drew the current "to the right" and you end up with a negative number, you know the (positive) current was really ...


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The currents will be the same. Using Kirchoff's current law, the total current going out of any node or any closed contour is 0. Suppose you have two resistances connected in parallel between nodes A and B. Draw a circle around the parallel resistances. The total current going out of the circle is 0. So outgoing current at A + outgoing current at B = 0. In ...


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To understand this better take the fluid mechanics analogy.Consider a pipe to be a conductor and the pressure across it to be the voltage difference and water flow to be current.Now consider a situation where there is water flowing from through the pipe,it is a matter of fact that if water is flowing there has to be some pressure difference between the ends ...


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You seem to be asking, "what happens to an electron when its voltage changes?" In short, nothing happens. An electron has no idea what voltage it's at. You claim this is different from an object's height, but it's actually exactly the same. If you only look at an object, by itself, you'll have no idea what height it's at. Nothing about the object changes ...


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So the spectral approach as I remember is you make an anszats $i=\sum_k a_k f_k(t)$ or something like that, you use some theorem to say that each component of the sum is independent so you get $a_k f_k''(t) + 5 a_k f_k'(t) + 1/4 a_k f_k(t)=0$, to get the factors (characteristic polynomial is what you said I believe) My guesses where the error could be: I ...


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If you did not have resistor R in the circuit the voltmeter would always give the same reading - the voltage of the battery. A thermistor changes its resistance when a temperature changes. You are not given a resistance meter. All you have is a battery, a resistor R and a voltmeter. The circuit as set up is called a potential divider which means that the ...


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In such circuit see that all resistance at rim have same potential, since no resistance reduced it. At rim all have same potential as that at $a$. At center all have same potential as that at $b$ So, all have same potential difference $V_{a} - V_{b}$ Same potential difference means that they are connected in parallel.


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The reason to use the alloy is because it has a much higher resistivity than copper and so the alloy wire will have a higher resistance which with standard laboratory apparatus can be measured more accurately. I also seem to remember that the temperature of coefficient is lower for some of these alloys ie for a given increase in temperature the resistance ...


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Short answer: "neither". A resistance, in a circuit, causes a voltage drop for a given current. Alternatively, if the voltage is given, it causes a particular current to flow. As soon as a circuit gets a little bit more complex (more than one battery-resistor-light bulb-switch), it becomes completely impractical to use lots of batteries to control ...



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